Mathematical Induction - Discrete Structures - Lecture Slides, Slides of Discrete Structures and Graph Theory

These solved exam paper are very easy to understand and very helpful to built a concept about the foundation of computers and discrete structures.The key points discuss in these notes are: Mathematical Induction, Natural Numbers, Propositional Function, Inductive Step, Positive Integers, Principle of Mathematical Induction, Product of Primes, Induction Hypothesis, Modus Ponens, Law of Detachment

Typology: Slides

2012/2013

Uploaded on 04/27/2013

ashakiran
ashakiran 🇮🇳

4.5

(27)

261 documents

1 / 14

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Follow me for a walk through...
Mathematical
Induction
Docsity.com
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

Partial preview of the text

Download Mathematical Induction - Discrete Structures - Lecture Slides and more Slides Discrete Structures and Graph Theory in PDF only on Docsity!

Follow me for a walk through...

• Mathematical

• Induction

  • The principle of mathematical induction is a useful tool for proving that a certain predicate is true for all natural numbers.
  • It cannot be used to discover theorems, but only to prove them.
  • Example:
  • Show that n < 2n^ for all positive integers n.
  • Let P(n) be the proposition “n < 2n^ .”
    1. Show that P(1) is true. (basis step)
  • P(1) is true, because 1 < 2^1 = 2.
    1. Show that if P(n) is true, then P(n + 1) is true. (inductive step)
  • Assume that n < 2n^ is true.
  • We need to show that P(n + 1) is true, i.e.
  • n + 1 < 2n+
  • We start from n < 2n^ :
  • n + 1 < 2n^ + 1 ≤ 2 n^ + 2 n^ = 2n+
  • Therefore, if n < 2n^ then n + 1 < 2n+
  • Another Example (“Gauss”):
  • 1 + 2 + … + n = n (n + 1)/
  1. Show that P(0) is true. (basis step)
  • For n = 0 we get 0 = 0. True.
  1. Show that if P(n) then P(n + 1) for any n∈N. (inductive step)
  • 1 + 2 + … + n = n (n + 1)/
  • 1 + 2 + … + n + (n + 1) = n (n + 1)/2 + (n + 1)
  • = (2n + 2 + n (n + 1))/
  • = (2n + 2 + n^2 + n)/
  • = (2 + 3n + n^2 )/
  • = (n + 1) (n + 2)/
  • = (n + 1) ((n + 1) + 1)/
  • There is another proof technique that is very similar to the principle of mathematical induction.
  • It is called the second principle of mathematical induction.
  • It can be used to prove that a propositional function P(n) is true for any natural number n.
  • The second principle of mathematical induction:
  • Show that P(0) is true. (basis step)
  • Show that if P(0) and P(1) and … and P(n), then P(n + 1) for any n∈N. (inductive step)
  • Then P(n) must be true for any n∈N. (conclusion)
  • Show that if P(2) and P(3) and … and P(n), then P(n + 1) for any n∈N. (inductive step)
  • Two possible cases:
  • If (n + 1) is prime , then obviously P(n + 1) is true.
  • If (n + 1) is composite , it can be written as the product of two integers a and b such that 2 ≤ a ≤ b < n + 1.
  • By the induction hypothesis , both a and b can be written as the product of primes.
  • Therefore, n + 1 = a⋅b can be written as the product of primes.
  • Then P(n) must be true for any n∈N. (conclusion)
  • End of proof.
  • We have shown that every integer greater than 1 can be written as the product of primes.